Problem: Graph this system of equations and solve. $6x+2y = 10$ $3x-2y = 8$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Answer: Convert the first equation, $6x+2y = 10$ , to slope-intercept form. $y = -3 x + 5$ The y-intercept for the first equation is $5$ , so the first line must pass through the point $(0, 5)$ The slope for the first equation is $-3$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move down (because it's negative) $1$ position to the right. $3$ positions down from $(0, 5)$ is $(1, 2)$ Graph the blue line so it passes through $(0, 5)$ and $(1, 2)$ Convert the second equation, $3x-2y = 8$ , to slope-intercept form. $y = \dfrac{3}{2} x - 4$ The y-intercept for the second equation is $-4$ , so the second line must pass through the point $(0, -4)$ The slope for the second equation is $\dfrac{3}{2}$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move up $2$ positions to the right. $3$ positions up from $(0, -4)$ is $(2, -1)$ Graph the green line so it passes through $(0, -4)$ and $(2, -1)$ The solution is the point where the two lines intersect. The lines intersect at $(2, -1)$.